Question

4yzp+q+2y=0

Answer 1

Solution:

Given,$\ 4 y z p+q+2 y=0$

Or,$\ 4 y z p+q=-2 y$ ...(1)

The Lagrange's auxiliary equations for (1) are

$\dfrac{d x}{4 y z}=\dfrac{d y}{1}=\dfrac{d z}{-2 y}$ ...(2)

Taking the first and third fractions of (2),

$d x+2 z d z=0$

On integrating,

$\Rightarrow x+z^{2}=c_{1}$ ...(3)

Taking the last two fractions of (2),

$d z+2 y d y=0$

On integrating,

$\Rightarrow z+y^{2}=c_{2}$ ...(4)

On adding (3) and (4), we get,

$\left(y^{2}+z^{2}\right)+(x+z)=c_{1}+c_{2}$

$\Rightarrow x+z^{2}+z+y^{2} =C$ [where $C=c_{1}+c_{2}$ ]

Hence, required answer is $x+z^{2}+z+y^{2} =C$